Two forces of magnitudes equal to 2p and p respectively act on a particle. If the first is doubled and second increased by 12 N, the direction of the resultant is unaltered. Then P =

Given that

${\text{F}}_{\text{1}} =\text{2P}$

${\text{F}}_{\text{2} }=\text{P}$

Then the direction of resultant R is $\alpha$in anticlockwise direction.

Then  $\tan \alpha =\frac{P\sin \theta }{2P+P\cos \theta }\to \left(1\right)$

Now F₁ is doubled i.e. ${{\text{F}}_1}^{|}=4\text{P}$ , F₂ is increased by 12

i.e. ${{\text{F}}_2}^{|}={\text{F}}_{\text{2}}+12$ now

also resultant ${\text{R}}^{\text{|}}$ is making same angle $\alpha$ with x – axis anticlockwise

direction.

$\tan \alpha =\frac{(P+12)\sin \theta }{4P+(P+12)\cos \theta }\to \left(2\right)$                            

We can have

$\Rightarrow \frac{P\sin \theta }{2P+P\cos \theta }=\frac{(P+12)\sin \theta }{4P+(P+12)\cos \theta }$                       

$\Rightarrow 4P^2+P^2\cos \theta +12P\cos \theta$

$\Rightarrow 2P^2+P^2\cos \theta +24P+12P\cos \theta$

$\Rightarrow {\text{2P}}^{\text{2}} =\text{ 24P}\Rightarrow \text{P}=\text{12 N}$